Question

Which of the following 1M solutions will have the highest concentration of ions?

A) Na₂[CdCl₄]
B) K₃[V(C₂O₄)₃]
C) K₂[MoOCl₄]
D) [Co(en)₂(C₂O₄)]Cl
E) [Zn(en)₂]Br₂

Answer 1

Here we have considered 1 M solution for all the complex compounds.

Hence each 1M solution will contain Avogadro’s number of complex compound.

How ions given by each complex compound will be different as discussed below.

(A) 1 Na₂[CdCl₄] -------> 2Na⁺ (aq) + [CdCl₄]^2- (aq)

    1 mol 2 mol          mol

Hence 1 mol of Na₂[CdCl₄] gives a total of 3 moles of ions.

(B) K₃[V(C₂O₄)₃] -------> 3K⁺ (aq) + [V(C₂O₄)₃]^3-(aq)

    1 mol                           3 mol             1 mol

Hence 1 mol of K₃[V(C₂O₄)₃] gives a total of 4 moles of ions.

(C) K₂[MoOCl₄] -----------> 2K⁺ (aq) + [MoOCl₄]^2-(aq)

           1 mol                            2 mol             1 mol

  

Hence 1 mol of K₂[MoOCl₄] gives a total of 3 moles of ions.

(D) ) [Co(en)₂(C₂O₄)]Cl --------à ) [Co(en)₂(C₂O₄)]⁺ (aq) + Cl^- (aq)

    1 mol                                    1 mol                            1 mol

Hence 1 mol of [Co(en)₂(C₂O₄)]Cl gives a total of 2 moles of ions.

(E)   [Zn(en)₂]Br₂ -----------à [Zn(en)₂]²⁺ (aq) + 2Br^- (aq)

1 mol                                  1 mol                  2 mol

Hence 1 mol of [Zn(en)₂]Br₂   gives a total of 3 moles of ions.

Hence among the above compounds (B) K₃[V(C₂O₄)₃] gives highest moles of ions.

Hece (B) is the correct answer.